How do you solve $abs(4x-3) = 5sqrt(x+4)$ ?

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Answer 1 · 2016-07-28T11:40:12

$x = 1/32 (49 pm 5 sqrt[329])$

$abs(4x-3) = 5sqrt(x+4)->(4x-3)^2=25abs(x+4)$

or

$16x^2-24x+9=16x^2-25x+x+4+5 = 25abs(x+4)$

or

$(16x^2-25x+5)/(x+4)+1=pm25$

we have two outcomes

$16x^2-25x+5=24(x+4)$ resulting in

$x = 1/32 (49 pm 5 sqrt[329])$

and

$16x^2-25x+5=-26(x+4)$ with

$x = 1/32 (-1 pm i sqrt[8511])$

the feasible solutions being

$x = 1/32 (49 pm 5 sqrt[329])$

How do you solve abs(4x-3) = 5sqrt(x+4)abs(4x-3) = 5sqrt(x+4)?